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At first order in the suppression of an effective operator, a factor of the leading Euler-Lagrange field equation defines a redundant operator removable by a local field redefinition. For ${\mathcal{L}}_0=(\partial \varphi {)}^2/2-m^2{\varphi }^2/2-\lambda {\varphi }^4/24$, one obtains ${\varphi }^3\square \varphi \doteq -m^2{\varphi }^4-\lambda {\varphi }^6/6$ and $(\square \varphi {)}^2\doteq m^4{\varphi }^2+m^2\lambda {\varphi }^4/3+{\lambda }^2{\varphi }^6/36$. Lower-dimensional parameter shifts must accompany this reduction when $m\ne 0$; it is not an identity for untransformed off shell correlation functions.

An off shell field configuration is not required to satisfy the Euler-Lagrange field equation. An external four-momentum is off shell if it is not constrained to its physical mass shell. On shell identities can therefore fail for general untransformed correlation functions, even when the associated S-matrix is unchanged by a field redefinition.

Route a directed ghost loop so that its internal four-momenta are $l$ and $l+k$. At the external gluon with indices $(\mu ,a)$, let the incoming ghost carry colour $c$ and the outgoing ghost carry colour $d$; the other Feynman vertex has the reverse order. The stated ghost-gluon Feynman rules supply $g_D^2{f}^{acd}{f}^{bdc}(l+k{)}^{\mu }{l}^{\nu }$. Their colour factor isThe two ghost propagators supply $i^2=-1$, while the closed ghost loop supplies another $-1$. Thus, with the insertion convention stated above,This establishes every prefactor before integration. To isolate its ultraviolet pole, take generic off shell $k$, initially $k^2<0$, so the massless integral has no overlapping infrared divergence. Introduce a Feynman parameter, shift $r=l+xk$, and set $\Delta =-x(1-x)k^2>0$. Odd powers of $r$ vanish, givingThe needed dimensional regularization integrals follow from the supplied gamma function formula. In $D=4-ϵ$,Here $\Gamma (ϵ/2)=2/ϵ+O(1)$, rather than $1/ϵ$. Using $1/D=1/4+O(ϵ)$ and ${\int }_0^{1}x(1-x)dx=1/6$, obtainTherefore the ghost-loop pole of the gluon self-energy isThe factor ${\mu }^{ϵ}$ in $g_D^2$ affects only finite terms at this order. The isolated ghost loop is not transverse: contracting this result gives $g^2{C}_A{\delta }^{ab}{k}^2{k}^{\nu }/(32{\pi }^2ϵ)$. Transversality is a condition on the complete gluon self-energy, including the other loop graphs.

Keep $m>0$ and set $p=p^{\prime }=0$, which is an off shell choice. The numerator isTerms odd in $k$ integrate to zero. Lorentz covariance and dimensional regularization give, inside this integral,Applying the gamma matrices identity from part (ii) once more yieldsConsequentlyThis massive zero-momentum QED vertex integral uses an integrated tensor identity for the replacement of $N^{\mu }$, not an equality for each fixed $k$. The physical photon remains massless. The retained fermion mass supplies a scale and makes this zero-external-four-momentum calculation suitable for extracting the ultraviolet pole without a scaleless integral.

There are four tree-level Feynman diagrams: three photon exchange graphs with pairings $12|34$, $13|24$, $14|23$, and the four-photon contact interaction. They are drawn in the bottom row of the figure. With no external fermions or Faddeev-Popov ghost fields, an internal matter or ghost line would have to close into a loop; it cannot add a tree-level Feynman diagram here.

Use all-incoming null four-momenta $k_i$ with ${\sum }_i{k}_i=0$. Let ${\epsilon }_i$ be the external photon polarization vectors, including the complex conjugation for physical outgoing photons, and impose $k_i\cdot {\epsilon }_i=0$. For one pairing $ij|kl$, put $K=k_i+k_j=-(k_k+k_l)$. At the first cubic Feynman vertex, the terms proportional to $k_i\cdot {\epsilon }_i$ and $k_j\cdot {\epsilon }_j$ vanish, leavingAt the other end, the incoming internal four-momentum is $+K$, giving $+2\lambda ({\epsilon }_k\cdot {\epsilon }_l)K^{\sigma }$. The Feynman gauge photon propagator then givesThe last expression is evaluated at generic kinematics, with singular channel limits taken after cancellation. Summing the three pairings produces $+4i{\lambda }^{2}P$, whereThe quartic Feynman vertex contributes $i{\mathcal{M}}_{\mathrm{contact}}=-4i{\lambda }^{2}P$. ThusThis is the tree photon-scattering cancellation in a quadratic Abelian gauge: the individual gauge fixing interactions cancel in the physical scattering amplitude. The argument uses transverse external photon polarization vectors; it does not assert that an arbitrary off shell four-point correlation function vanishes. Physical light-by-light scattering first occurs through charged-particle loop Feynman diagrams.

The superficial degree of divergence of this scalar electrodynamics three-point vertex is one. Its ultraviolet pole is consequently a local expression at most linear in the external four-momenta. The allowed counterterm Lagrangian, with the charge-conjugation and Hermiticity properties of scalar electrodynamics, contains a single one-derivative scalar electrodynamics three-point vertex:Thus the coefficient ${\delta }_1$ is independent of the chosen external four-momenta. Setting $k^{\prime }=0$ leaves $-ie{\delta }_1{k}^{\mu }$, so a nonzero generic $k$ still determines that coefficient. It does not determine the full finite scattering amplitude at other four-momenta.

Keep $m\ne 0$ and generic off shell $k$ while extracting the ultraviolet pole. These choices avoid spurious infrared divergences; setting all scales to zero would instead produce scaleless integrals that can conceal the ultraviolet pole.

The scalar QED Ward identity equates the vertex renormalization and wave-function renormalization constants. Consequently, using part (iv),The renormalization constants describe gauge-dependent fields and off shell vertices, so their dependence on $\xi$ is permitted. A physical scattering amplitude includes all Feynman diagrams, the counterterms, and the external-state factors from LSZ reduction. The Ward identities ensure that the dependence on $\xi$ cancels for physical external states. Infrared-sensitive probabilities also require the usual inclusive treatment of unobserved soft photons. Thus no dependence on $\xi$ is implied for consistently calculated physical observables.

There are exactly four tree-level Feynman diagrams: one four-photon contact graph and one exchange graph for each of the three partitions $12|34$, $13|24$ and $14|23$ of the external legs. The exchange graphs contain two cubic Feynman vertices. A fermion loop or a closed Faddeev-Popov ghost line has positive loop order, so neither enters here.

Write $p_1+p_2+p_3+p_4=0$ and take physical photon polarization vectors ${\epsilon }_i$ with $p_i\cdot {\epsilon }_i=0$ and $p_i^2=0$. At the cubic Feynman vertex attached to legs $i,j$, the two terms containing $p_i\cdot {\epsilon }_i$ or $p_j\cdot {\epsilon }_j$ vanish. If $q=p_i+p_j$ flows into the internal line, the remaining factor at that end isAt the other end the internal incoming four-momentum is $q$ rather than $-q$, giving $+2\lambda q^{\nu }({\epsilon }_k\cdot {\epsilon }_l)$. Multiplication by the photon propagator givesThe cancellation holds first at generic nonzero channel Mandelstam invariants and then by continuation at degenerate kinematics. Summing the three exchange graphs produces $4i{\lambda }^{2}P$, whereThe quartic Feynman vertex contributes $-4i{\lambda }^{2}P$. ThereforeThis is a cancellation for physical on shell photon polarization vectors. The off shell four-photon correlation function need not vanish. Physical light-by-light scattering first occurs through charged-particle loop Feynman diagrams.

At large loop momentum, each Dirac propagator scales as ${\ell }^{-1}$. One four-dimensional loop and four internal Dirac propagators therefore give the superficial degree of divergenceThus a possible ultraviolet divergence in the sum is logarithmic. By locality its divergent coefficient is a degree-zero polynomial in the external four-momenta, namely a constant Lorentz tensor,The photon Ward identities apply to the complete sum for arbitrary external four-momenta, including off shell ones: $k_{1\mu }{\mathcal{M}}^{\mu \nu \rho \sigma }=0$, and likewise for the other legs. A constant tensor with $k_{1\mu }{\mathcal{M}}_{\mathrm{div}}^{\mu \nu \rho \sigma }=0$ for every $k_1$ must be zero. Hence $a=b=c=0$ andEquivalently, a logarithmic divergence would require a local dimension-four $A^4$ counterterm, forbidden by Abelian gauge invariance. A gauge-invariant four-photon local operator starts with four gauge field strengths, for example $(F_{\mu \nu }{F}^{\mu \nu }{)}^2$, of mass dimension eight; it cannot be this power counting divergence. Individual boxes need not be finite; their forbidden divergent parts cancel in the sum. There are no divergent proper loop subgraphs at this loop order. With the massive Dirac field in the source, there is also no soft or collinear infrared divergence at generic external kinematics. The argument establishes ultraviolet finiteness, and does not claim the absence of physical threshold singularities.

Use the sign convention $E(\varphi )=\delta S_0/\delta \varphi$ from part b, and letIn the path integral, make the local field redefinition ${\varphi }_{\mathrm{old}}=\chi -\eta f(\chi )$. The functional Taylor expansion of the action givesThe two first-order terms cancel. Neglecting the functional Jacobian as permitted,Thus $fE$ is a redundant operator at first order, by the equivalence theorem for field redefinitions. The same argument works for a local derivative-dependent $f$, which is needed to remove $(\square \varphi {)}^2$. Insertions and sources must be transformed too: arbitrary untransformed off shell correlation functions need not be equal.

Sign qualification. If instead $E_{+}=\square \varphi +m^2\varphi +\lambda {\varphi }^3/6$ is used, then $E_{+}=-\delta S_0/\delta \varphi$. An added $+\eta f{E}_{+}$ is removed by ${\varphi }_{\mathrm{old}}=\chi +\eta f(\chi )$. The minus shift requested in the source therefore fixes which sign is meant by its otherwise unspecified $E$.

Let $a=\lambda /6$. Within terms already suppressed by ${\Lambda }^{-2}$, the leading equation of motion gives $\square \varphi =-m^2\varphi -a{\varphi }^3$. Equality here means equivalence under a perturbative field redefinition, together with integration by parts, rather than a pointwise identity inside arbitrary off shell correlation functions. HenceFor the second relation, factor the difference as $(\square \varphi -m^2\varphi -a{\varphi }^3)(\square \varphi +m^2\varphi +a{\varphi }^3)$, which is explicitly a redundant operator. Ifthe reduced correction isThe first two coefficients shift the already present mass and quartic coupling constant. After these shifts,This is the equation-of-motion reduction of scalar operators through $O({\Lambda }^{-2})$; terms of $O({\Lambda }^{-4})$ are outside that truncation. The lower-dimensional shifts should not be silently dropped when $m\ne 0$.

In the right panel of part a, $p$ enters the fermion line, $p^{\prime }=p+q$ leaves it, and a virtual photon of momentum $k$ bridges the external photon vertex. Write ${\Gamma }^{\mu }={\gamma }^{\mu }+{\Lambda }^{\mu }$ and $F_1=1+F_1^{(1)}+\cdots$. The one-loop amplitude isThis Feynman integral has a logarithmic superficial degree of divergence. Its ultraviolet pole is independent of the external momenta and $m$, because every such numerator insertion or denominator correction lowers the large-$k$ degree. Preserve an auxiliary positive mass $M$ while extracting the ultraviolet pole; setting everything to zero would create a scaleless integral and conceal the cancellation of ultraviolet and infrared divergences.

The leading numerator is ${\gamma }^{\alpha }/k{\gamma }^{\mu }/k{\gamma }_{\alpha }$. Symmetric integration replaces $k_{\rho }{k}_{\sigma }$ by $g_{\rho \sigma }{k}^2/d$. Two contractions of gamma matrices then giveFor the auxiliary-mass extraction of ultraviolet poles,The second term is ultraviolet convergent. ThusThe ultraviolet pole of the QED vertex is therefore ${\Lambda }^{\mu }{|}_{\mathrm{UV}}=e^2{\gamma }^{\mu }/(16{\pi }^2ϵ)$. The Pauli structure has zero trace against ${\gamma }_{\mu }$ and no ultraviolet pole at this order.

Form-factor qualification. The two electron electromagnetic vertex form factors in the source describe matrix elements between external on shell Dirac spinors. A general off shell vertex has more structures, so the unprojected trace is not a projector for the complete finite on-shell $F_1$. It does correctly isolate the divergent local ${\gamma }^{\mu }$ coefficient, which is all that is required here.

First take the trace in four spacetime dimensions:The product rule gives $\square {\varphi }^2=2(\partial \varphi {)}^2+2\varphi \square \varphi$. For the massless equation of motion, $\varphi \square \varphi =-\lambda {\varphi }^4/6$, soBoth coefficients vanish forThe minus sign follows from the order ${\partial }^{\mu }{\partial }^{\nu }-{\eta }^{\mu \nu }\square$ used in the question's stress-energy tensor improvement. Commuting partial derivatives shows that this improvement has identically zero divergence, so it preserves the conservation law.

The precise trace identity, without using the equation of motion, isHence the massless tensor is traceless on shell. It is not traceless for arbitrary off shell configurations: a nonzero constant $\varphi =A$ with $\lambda \ne 0$ has ${\Omega }^{\mu }{}_{\mu }=\lambda A^4/6$. This is a classical result; it does not assert absence of a quantum trace anomaly.

The nonlinear gauge fixing $\partial \cdot A+\lambda A^2$ produces cubic and quartic photon Feynman vertices. For physical photon polarization vectors, the two-external-leg cubic vertex reduces to a longitudinal internal momentum times the scalar product of the external polarizations. Its exchanged photon propagator cancels that momentum squared. Each of the three exchange graphs contributes $+4i{\lambda }^2({\epsilon }_i\cdot {\epsilon }_j)({\epsilon }_k\cdot {\epsilon }_l)$, cancelling the corresponding term of the quartic vertex. This vanishing physical tree-level Feynman diagram sum does not assert that the off shell four-point function vanishes.

Write the QED vertex as $-ie({\gamma }^{\mu }+{\Lambda }^{\mu })$. In Feynman gauge and $d=4-2ϵ$, the one-loop ultraviolet pole has the local form ${\Lambda }_{\mathrm{UV}}^{\mu }=e^2{\gamma }^{\mu }/(16{\pi }^2ϵ)$. Its coefficient follows from the large-momentum gamma matrices contraction ${\gamma }^{\alpha }/k{\gamma }^{\mu }/k{\gamma }_{\alpha }\mapsto (d-2{)}^2{k}^2{\gamma }^{\mu }/d$. It is independent of external kinematics, whereas the full finite vertex requires additional structures off shell. Its cancellation verifies the Ward identity for QED renormalization constants.